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In Wolfgang Nolting's 'Analytical Mechanics', the concept of 'generalized potential' is discussed:

For non-conservative systems, but with holonomic constraints, instead of that, the starting point must be : $$ \frac{d}{d t} \frac{\partial T}{\partial \dot{q}_j}-\frac{\partial T}{\partial q_j}=Q_j, \quad j=1,2, \ldots, S . $$

However, we come to formally unchanged Lagrange equations for the so-called Generalized potentials $$ U=U\left(q_1, \ldots, q_S, \dot{q}_1, \ldots, \dot{q}_S, t\right) $$ if the generalized forces $Q_j$ are derivable from $U$ by: $$ Q_j=\frac{d}{d t} \frac{\partial U}{\partial \dot{q}_j}-\frac{\partial U}{\partial q_j}, \quad j=1,2, \ldots, S . $$

The first term on the right-hand side is new compared to the case of a conservative system. For the Generalized Lagrangian function $$ L=T-U $$

Suppose we have the equation $\frac{d}{d t} \frac{\partial T}{\partial \dot{q}_j}-\frac{\partial T}{\partial q_j}=Q_j$, where $j=1,2, \ldots, S$, and $T$ is explicitly given, say, as a simple quadratic equation. Since the equation involving the generalized potential $U$ is consistent with it, expressed as $\frac{d}{d t} \frac{\partial U}{\partial \dot{q}_j}-\frac{\partial U}{\partial q_j}$ for $j=1,2, \ldots, S$, can I directly consider $T$ as $U$ by this method to determine $U$? Or $T$ and $U$ differ by a constant. Would such a definition of $U$ not cause confusion?

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guoxu
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1 Answers1

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Comments to the question (v2):

  1. The generalized velocity-dependent potential $U$ exists in many important cases, but does not exist for e.g. dissipative forces, cf. e.g. this Phys.SE post.

  2. In a nutshell, the Lagrange equations ("EOMs") equate kinematics ("acceleration") with dynamics ("forces").

  3. The kinetic term $T$ and the dynamic term $U$ should not be conflated. Doing so may turn a useful non-trivial EOM into a useless trivial identity.

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